Question

${CN}^{-}⟶CN{O}^{-}$
The number of electrons that must be added in balancing the half reaction is:

• A. $0$
• B. $1$ on the right
• C. $1$ on the left
• D. $2$ on the right

Answer 1

Answer: (D)

$2$ on the right

${CN}^{-}⟶CN{O}^{-}$
Let the oxidation number of $C$ in $C{N}^{-}$ be $x$ and in $CN{O}^{-}$ be $y$.
$x-3=-1;x=+2$
$y-3-2=-1;y=+4$
So, the oxidation state of $C$ changes by $+2$.
${CN}^{-}⟶CN{O}^{-}+2e$